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The cross-sectional area of the plane shown in the figure is equal to 1 cm². A current of 2 A flows through the conductor. The current density at point P in the conductor will be ______.

Asked in RS Academy GUJCET booklet · Drift velocity, mobility and current density

• (1) 4/(√3)×10⁴ A m⁻²
• (2) (√3)/2×10⁴ A m⁻²
• (3) 3/2×10⁴ A m⁻²
• (4) (√3)/4×10⁴ A m⁻²

Step-by-step solution

Why the other options are wrong

• (2) No single slip produces this value, and it is impossible anyway: the tilted plane is at least as large as the true cross-section, so J ≥ (2 A)/(10⁻⁴ m²) = 2×10⁴ A m⁻².
• (3) No single slip produces this value. Sanity check: J = I/(A cos 30°) = 2.31×10⁴ A m⁻², and nothing below 2×10⁴ A m⁻² is possible because the tilted plane is larger than the true cross-section.
• (4) No single slip produces this value. At about 0.43×10⁴ A m⁻² it is below the floor of 2×10⁴ A m⁻² set by I/(1 cm²), which the true (smaller) cross-section can only raise.

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